Giải hộ mình câu này với các bạn

LHTDEMON

a)

$A\left(x\right)=7x^3-9-x^2+\frac{1}{3}x+2x^2-7x^3-\frac{2}{3}x$

$=\left(7x^3-7x^3\right)+\left(-x^2+2x^2\right)+\left(\frac{1}{3}x-\frac{2}{3}x\right)-9$

$=x^2-\frac{1}{3}x-9$

$B\left(x\right)=-7x^3+x^2-x+3+\frac{1}{3}x+2-4x^2$

$=-7x^3+\left(x^2-4x^2\right)+\left(-x+\frac{1}{3}x\right)+\left(3+2\right)$

$=-7x^3+3x^2-\frac{2}{3}x+5$

b)

$A\left(x\right)+B\left(x\right)=\left(x^2-\frac{1}{3}x-9\right)+\left(-7x^3-3x^2-\frac{2}{3}x+5\right)$

$=-7x^3+\left(x^2-3x^2\right)+\left(-\frac{1}{3}x-\frac{2}{3}x\right)+\left(-9+5\right)$

$=-7x^3-2x^2-x-4$

$A\left(x\right)-B\left(x\right)=\left(x^2-\frac{1}{3}x-9\right)-\left(-7x^3-3x^2-\frac{2}{3}x+5\right)$

$=7x^3+\left(x^2+3x^2\right)+\left(-\frac{1}{3}x+\frac{2}{3}x\right)+\left(-9-5\right)$

$=7x^3+4x^2+\frac{1}{3}x-14$

$B\left(x\right)-A\left(x\right)=\left(-7x^3+3x^2-\frac{2}{3}x+5\right)-\left(x^2-\frac{1}{3}x-9\right)$

$=-7x^3+\left(3x^2-x^2\right)+\left(-\frac{2}{3}x+\frac{1}{3}x\right)+\left(5+9\right)$

$=-7x^3+2x^2-\frac{1}{3}x+14$

c)

$H\left(x\right)=-7x^3-2x^2-x-4$

Hệ số cao nhất: -7

Hệ số tự do: -4

d)

$H\left(0\right)=-7.0^3-2.0^2-0-4=-4$

$H\left(-1\right)=-7.\left(-1\right)^3-2.\left(-1\right)^2-\left(-1\right)-4=7-2+1-4=2$

$H\left(\frac{1}{2}\right)=-7.\left(\frac{1}{2}\right)^3-2.\left(\frac{1}{2}\right)^2-\frac{1}{2}-4=-7.\frac{1}{8}-2.\frac{1}{4}-\frac{1}{2}-4=-\frac{47}{8}$

e)

$H\left(x\right)+4=0$

$-7x^3-2x^2-x-4+4=0$

$-7x^3-2x^2-x=0$

$x\left(-7x^2-2x-1\right)=0$

Trường hợp 1: $x=0$

Trường hợp 2: $-7x^2-2x-1=0$

$7x^2+2x+1=0$

Ta có:

$7\left(x^2+\frac{2}{7}x\right)+1=7\left\lbrack\left(x+\frac{1}{7}\right)^2-\frac{1}{49}\right\rbrack+1=7\left(x+\frac{1}{7}\right)^2-\frac{1}{7}+1=7\left(x+\frac{1}{7}\right)^2+\frac{6}{7}>0$ với mọi x

$\Rightarrow$ Phương trình trên vô nghiệm

Vậy x = 0.